悬空指针示例

Question

在下面的代码中,为什么会s1.printVal导致悬空指针错误？是不是s1直到它的销毁的对象,即它的指针,仍然可以访问？

class Sample
{
  public:
    int *ptr;
    Sample(int i)
    {
        ptr = new int(i);
    }

    ~Sample()
    {
        delete ptr;
    }
    void PrintVal()
    {
        cout << "The value is " << *ptr;
    }
};

void SomeFunc(Sample x)
{
    cout << "Say i am in someFunc " << endl;
}

int main()
{
    Sample s1 = 10;
    SomeFunc(s1);
    s1.PrintVal(); // dangling pointer
}

资源

Answer 1

这里的问题是为参数做的副本SomeFunc().该副本在销毁时取消分配指针.您还需要实现复制构造函数和复制赋值运算符.见三规则.

编辑:

这是"扩展"的伪代码,即编译器在main()函数中为您做的事情:

// main
addr0 = grab_stack_space( sizeof( Sample )); // alloc stack space for s1
Sample::ctor( addr0, 10 );                   // call ctor of Sample
addr1 = grab_stack_space( sizeof( Sample )); // alloc stack for argument
Sample::ctor( addr1, addr0 );                // call COPY-ctor of Sample
SomeFunc( addr1 );                           // call SomeFunc
Sample::dtor( addr1 );                       // XXX: destruct the copy
free_stack_space( addr1, sizeof( Sample ));  // free stack taken by copy
Sample::PrintVal( addr0 );                   // call member func on s1
Sample::dtor( addr0 );                       // destruct s1
free_stack_space( addr0, sizeof( Sample ));  // YYY: free stack taken by s1

这不是确切的表示,而是概念性的解释.只需考虑编译器与代码有什么关系.

指针成员在标记为的步骤处Sample为delete-ed XXX,然后delete在该步骤再次执行-ed YYY.